众所周知，目前的mysql版本中并不支持直接的递归查询，但是通过递归到迭代转化的思路，还是可以在一句SQL内实现树的递归查询的。这个得益于Mysql允许在SQL语句内使用@变量。以下是示例代码。

创建表格

CREATE TABLE `treenodes` ( `id` int , -- 节点ID `nodename` varchar (60), -- 节点名称 `pid` int -- 节点父ID );

插入测试数据

INSERT INTO `treenodes` (`id`, `nodename`, `pid`) VALUES ('1','A','0'),('2','B','1'),('3','C','1'), ('4','D','2'),('5','E','2'),('6','F','3'), ('7','G','6'),('8','H','0'),('9','I','8'), ('10','J','8'),('11','K','8'),('12','L','9'), ('13','M','9'),('14','N','12'),('15','O','12'), ('16','P','15'),('17','Q','15'),('18','R','3'), ('19','S','2'),('20','T','6'),('21','U','8');

查询语句

SELECT id AS ID,pid AS 父ID ,levels AS 父到子之间级数, paths AS 父到子路径 FROM ( SELECT id,pid, @le:= IF (pid = 0 ,0, IF( LOCATE( CONCAT('|',pid,':'),@pathlevel) > 0 , SUBSTRING_INDEX( SUBSTRING_INDEX(@pathlevel,CONCAT('|',pid,':'),-1),'|',1) +1 ,@le+1) ) levels , @pathlevel:= CONCAT(@pathlevel,'|',id,':', @le ,'|') pathlevel , @pathnodes:= IF( pid =0,',0', CONCAT_WS(',', IF( LOCATE( CONCAT('|',pid,':'),@pathall) > 0 , SUBSTRING_INDEX( SUBSTRING_INDEX(@pathall,CONCAT('|',pid,':'),-1),'|',1) ,@pathnodes ) ,pid ) )paths ,@pathall:=CONCAT(@pathall,'|',id,':', @pathnodes ,'|') pathall FROM treenodes, (SELECT @le:=0,@pathlevel:='', @pathall:='',@pathnodes:='') vv ORDER BY pid,id ) src ORDER BY id

最后的结果如下：

ID 父ID父到子之间级数 父到子路径

------ ------ ---------------------------------------

1 0 0 ,0

2 1 1 ,0,1

3 1 1 ,0,1

4 2 2 ,0,1,2

5 2 2 ,0,1,2

6 3 2 ,0,1,3

7 6 3 ,0,1,3,6

8 0 0 ,0

9 8 1,0,8

10 8 1,0,8

11 8 1 ,0,8

12 9 2,0,8,9

13 9 2 ,0,8,9

14 12 3 ,0,8,9,12

15 12 3 ,0,8,9,12

16 15 4 ,0,8,9,12,15

17 15 4 ,0,8,9,12,15

18 3 2 ,0,1,3

19 2 2 ,0,1,2

20 6 3 ,0,1,3,6

21 8 1 ,0,8

以上就是一句SQL实现MYSQL的递归查询的实现全过程，希望对大家的学习有所帮助。